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Let V be a $\mathbb{Z}[i]$-module generated by elements $v_1$ and $v_2$ with the following relations:

$$ (1+i)v_1+(2-i)v_2=0 $$ and $$ 3v_1+5iv_2=0 $$

I need to write V as a sum of cyclic modules. I tried to do this the basic way: putting the matrix $$ \left[\begin{array}{cc}1+i&2-i\\3&5i \end{array} \right] $$ in the Smith form $$ \left[\begin{array}{cc}1&0\\0&3-19i \end{array} \right] $$ Now, by the Structure Theorem, we have that $V$ is isomorphic to $\frac{\mathbb{Z}[i]}{\langle 3-19i\rangle}$, which is cyclic because $\langle 3-19i\rangle$ is an ideal of $\mathbb{Z}[i]$. Therefore nothing more needs to be done.

Is that correct?

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  • $\begingroup$ Dear Gustavo, I think there is an error in your computations. E.g. the determinant of the original matrix is $-11 + 8i$, which is not the same as $3 - 19i$ (even up to a unit). Regards, $\endgroup$ – Matt E Jul 4 '12 at 4:01
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    $\begingroup$ Ok, I'm going to make my calculations again. Regardless, can you tell me if the idea is correct? $\endgroup$ – Marra Jul 4 '12 at 5:11
  • $\begingroup$ Dear Gustavo, Yes, the idea is correct. Regards, $\endgroup$ – Matt E Jul 4 '12 at 13:25
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    $\begingroup$ I made my calculations again, I noticed that in one particular step I was multiplying a row by a factor which was not an unit (in $\mathbb{Z}[i]$); now the result is the matrix $$\left[\begin{array}{cc}1&0\\0&-11+8i\end{array}\right]$$ just as you pointed through the determinant. Thus the result is that V is isomorphic to $\frac{\mathbb{Z}[1]}{<-11+8i>}$. Thanks! $\endgroup$ – Marra Jul 4 '12 at 20:12
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    $\begingroup$ Dear Gustavo, You're welcome. I am writing an answer reflecting our exchange in the comments, just so that this question can move from the unanswered to the answered category. Best wishes, $\endgroup$ – Matt E Jul 5 '12 at 2:52
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Here is an answer, based on the discussion in the comments:

The method is correct, but there was a calculation mistake. The correct answer is that the module $V$ is isomorphic to $\mathbb Z[i]/\langle -11 + 8i \rangle.$

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